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Lunna [17]
1 year ago
9

3. Classify each function according to whether it is a vertical stretch, a vertical compression, a horizontal

Mathematics
1 answer:
Vladimir79 [104]1 year ago
4 0

The classifications of the functions are

  • A vertical stretch --- p(x) = 4f(x)
  • A vertical compression --- g(x) = 0.65f(x)
  • A horizontal stretch --- k(x) = f(0.5x)
  • A horizontal compression  --- h(x) = f(14x)

<h3>How to classify each function accordingly?</h3>

The categories of the functions are given as

  • A vertical stretch
  • A vertical compression
  • A horizontal stretch
  • A horizontal compression

The general rules of the above definitions are:

  • A vertical stretch --- g(x) = a f(x) if |a| > 1
  • A vertical compression --- g(x) = a f(x) if 0 < |a| < 1
  • A horizontal stretch --- g(x) = f(bx) if 0 < |b| < 1
  • A horizontal compression  --- g(x) = f(bx) if |b| > 1

Using the above rules and highlights, we have the classifications of the functions to be

  • A vertical stretch --- p(x) = 4f(x)
  • A vertical compression --- g(x) = 0.65f(x)
  • A horizontal stretch --- k(x) = f(0.5x)
  • A horizontal compression  --- h(x) = f(14x)

Read more about transformation at

brainly.com/question/1548871

#SPJ1

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Leno4ka [110]

Answer:

Firstly need to understand the questions.

If 15 men working 10 days, They earn 500.

If 12 men working 14 days, how much they will earn?

Step-by-step explanation:

560

Solution:

15 men working for 10 days =500

1 man working for 1 day = 500/(10*15) =10/3

Now 12 men working for 14 days = 12*14*10/3 =560.

I am pretty sure that answer is right.

Thank you :)

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do you know the length of each side of the parallelogram?

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Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt
ser-zykov [4K]

Answer:

I(t)=\frac{1}{3}(1-e^{-30t})

Step-by-step explanation:

We are given that

\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}

R=150 ohm

L=5 H

V(t)=10 V

P=\frac{R}{L}=\frac{150}{5}=30

I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}

I(t)\times I.F=\int e^{30 t}\times 10 dt+C

I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C

I(t)=\frac{1}{3}+Ce^{-30 t}

I(0)=0

Substitute t=0

0=\frac{1}{3}+C

C=-\frac{1}{3}

Substitute the values

I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}

I(t)=\frac{1}{3}(1-e^{-30t})

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Describe the end behavior of the following function: F(x)=2x^4+x^3
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Answer:

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Since, the given function is f(x)=2x^{4}+x^{3}, and the end behavior of the given function is determined as:

Consider the given function f(x)=2x^{4}+x^{3}, identify the degree of the function:

The degree of the function is : 4 which is even

And then identify the leading coefficient of the given function that is +2 which is positive in nature.

Hence, the function is positive and even in nature, therefore, the end behavior of the function will be rising to the left and rising to the right.

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